Question

我无法解释为什么我的查询会返回错误。我用PHP生成它。

$column = "display_name";
$criteria = "Steven";
$query = "
                SELECT
                    m.user_id AS id, m.email_address, m.display_name, m.status, UNIX_TIMESTAMP(m.join_date) AS join_date,
                    l.listing_id, COUNT(l.member_id) AS total_listings,
                    g.group_id AS group_id, g.title AS group_title
                FROM users AS m
                LEFT JOIN listings as l on m.user_id = l.member_id
                LEFT JOIN groups AS g on m.group_id = g.group_id";

            if($column == "display_name"){

                $query .= '
                WHERE m.display_name LIKE \'%'.$criteria.'\'%';

            } else {

                $query = "
                WHERE m.".$column." = '$criteria'";

            }

            $query .="
                GROUP BY m.user_id";

以上代码产生以下内容：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '% GROUP BY m.user_id' at line 8

SELECT m.user_id AS id, m.email_address, m.display_name, m.status, UNIX_TIMESTAMP(m.join_date) AS join_date, l.listing_id, COUNT(l.member_id) AS total_listings, g.group_id AS group_id, g.title AS group_title FROM users AS m LEFT JOIN listings as l on m.user_id = l.member_id LEFT JOIN groups AS g on m.group_id = g.group_id WHERE m.display_name LIKE '%Steven'% GROUP BY m.user_id

我尝试过使用LIKE %$criteria%和LIKE '%$criteria%'，我不知道为什么它们都不起作用。有什么想法吗？

Answer 1

您的百分比超出了引号：

$query .= '
WHERE m.display_name LIKE \'%'.$criteria.'%\'';

使用双引号可以使事情变得更清晰：

$query .= "
WHERE m.display_name LIKE '%$criteria%'";

我当然希望你能够很好地逃避这些数据。

Answer 2

因为您的查询正在被解释为

WHERE m.display_name LIKE '%text'%

在这里，您在%签名之前使用单引号。它应该是'%text%'而不是'%text'%。像这样修复它

$query .= '
            WHERE m.display_name LIKE \'%'.$criteria.'%\'';

最好在外面使用双引号来消除所有这些混淆。像这样：

$query .= "
            WHERE m.display_name LIKE '%$criteria%'";

为什么我的MySQL查询返回语法错误？

2 个答案: